SELECTED DISCUSSIONS TO exercises/problems
BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE
DESIGNED TO HELP YOU DO RELATED EXERCISES.

error log #1 --See corrections to #6 below regarding the
forces perpendicular to the incline.
error log #2 --See corrections to #26 below regarding the leftward
motion in which the car slows down; the cable would swing left in
that case.

5. SEE EXAMPLE 5.2 FOR A GEOMETRICAL 2D (2 dimensional)
system. Resolve the vector components along horizontal and
vertical axes.

6. See example 5.3 for a simplified version of problem; This time, in
contrast to the previous problem (#5), we resolve our components along x
and y axes that are parallel and perpendicular to the inclined plane, respectively.
Let the positive x direction be UP the incline.
SUM of forces in x direction = pos - neg = 0 = T*cos 31 -
mg*sin 25.
SUM of forces in y direction = pos - neg = 0 = N + T*sin31 - mg*cos
25.

Use these equations to answer parts (b) and (c) to find T AND
N; use the definition of normal force, exerted by the surface on the
car.

23. See example 5.5 for the spirit of
this problem.
Let ay be the y-component of acceleration .

(a) Let up be the positive y direction, the assumed direction of motion.
sum of forces in y direction = m*ay = pos - neg = Fs
- mg, where Fs is the magnitude of the spring force on the
fish; note the value Fs is the scale reading ! The scale
reading is Fs = 60.0 (N).
Now, you are given ay = + 2.45 m/s2.
Solve for m and find mg.
(b) Let DOWN be the positive y direction, the assumed direction of
motion. In the previous part, we we see that mg < 60.0
(N). This must be true otherwise the fish would not be speeding up
as it moves upward. If the fish accelerates downward,
i.e., speeds up as it moves down, we have a different situation. In that
case, mg > Fs. Assuming the object moves down, we have
sum of forces in y direction = m*ay = pos - neg =
mg - Fs . You are given Fs
= 35.0 (N). Solve for the acceleration
component.
(c) ZERO in which case the fish would be in free fall.

26. See example 5.5 for a direct
application of this problem to a real world situation. As the car moves
right, speeding up, the acceleration points right and the cable swings left.
Meanwhile, as the car moves left, the cable would swing right if
the speed was increasing. But the speed decreases during leftward
motion, which means the acceleration points right and thus the cable
swings left as in part (a) .
Note in part (c) the cable will hang vertically with no deflection
since the acceleration is ZERO.

28. See class notes on #29 we worked
on the board. Let a
be the magnitude of the common acceleration of the boxes.

39. (a) sum of forces in x direction = pos
= neg = 0 since the velocity is constant as the box moves along the
horizontal.
0 = F - fk, where F is the magnitude of person's push
force and fk is the friction force magnitude. Note: = uk*N,
where uk is the coefficient of kinetic friction and N is the normal
force magnitude.

sum of the forces in the y direction = pos - neg = 0 = N - mg;
thus N = mg.
Find F.
(b) Yo, in this case the only horizontal force acting on the
slowing down box is kinetic friction and the ACCELERATION IS NOT
ZERO.

From part (a), N = mg as before, but now we have a different
situation in the x-direction:

Going up the incline, sum of forces in the x direction = m* ax
= pos - neg = 0 - fk - mg*sin 40 since all the
x-directed -force components point down incline; there is no
"pos" force along x axis ! Note: = uk*N,
where uk is the coefficient of kinetic friction and N is the normal
force magnitude.
Note: sum of forces in y direction = pos - neg = N - mg*cos 40; thus N =
mg*cos40.
Find the acceleration.
(b) Let down the incline be the positive x direction. Going down the
incline, sum of forces in the x direction =
m* ax = pos - neg =mg*sin 40 - fk
since the x-directed-weight component points down
incline and friction points in the opposite direction (up). Note:
= uk*N, where uk is the coefficient of kinetic
friction and N is the normal force magnitude.
Note: sum of forces in y direction = pos - neg = N - mg*cos 40; thus N =
mg*cos40.
Find the acceleration. NOTE: IF we chose to keep up the incline as
positive, the acceleration would have the same magnitude but opposite
sign.

52. See example 5.13: Let down be
positive; sum of forces in y direction =
m*a = mg - FD, where a is the
magnitude of the downward acceleration and FD is
the drag force magnitude. In the formula for FD
set vy = vT / 2 , where the terminal
velocity vT is derived in the example.